Problem on density of set generated by irrational number

Question

This is problem 2.12 from Sohrab's Basic Real Analysis.

Problem

Let $\alpha>0$ be irrational. Show that for each $n\in\mathbb{Z}$ there is a unique integer $k_n\in\mathbb{Z}$ such that $k_n\alpha\leq n<(k_n+1)\alpha$. Let $x_n:=n-k_n\alpha$. Show that $\{x_n\mid n\in\mathbb{Z}\}$ is dense in $[0,\alpha]$.

Proof (incomplete)

First we prove that for each $n\in\mathbb{Z}$ there is a unique integer $k_n\in\mathbb{Z}$ such that $k_n\alpha\leq n<(k_n+1)\alpha$: If there were two such integers, $k_n,k_n'$, then $$ k_n\leq\frac{n}{\alpha}<k_n+1,\quad k_n'\leq\frac{n}{\alpha}<k_n'+1. $$ But for $k_n\neq k_n'$, $n/\alpha$ cannot be in both disjoint intervals $[k_n,k_n+1)$ and $[k_n',k_n'+1)$, leading to a contradiction.

Now we define $x_n:=n-k_n\alpha$ and $A:=\{x_n\mid n\in\mathbb{Z}\}$. We prove that $A$ is dense in $[0,\alpha]$. From the first result we know that $0\leq x_n<\alpha$.

Question

How do I prove the second part?

I proved the following result (Problem 2.11(c)): Deduce from part (a) that the set $\{n\alpha-[n\alpha]\mid n\in\mathbb{Z}\}$ is dense in $[0,1]$.

But I am not able to connect the two things.

Answer 1

Thank to i707107, here the completed proof:

Proof

First we prove that for each $n\in\mathbb{Z}$ there is a unique integer $k_n\in\mathbb{Z}$ such that $k_n\alpha\leq n<(k_n+1)\alpha$: If there were two such integers, $k_n,k_n'$, then $$ k_n\leq\frac{n}{\alpha}<k_n+1,\quad k_n'\leq\frac{n}{\alpha}<k_n'+1. $$ But for $k_n\neq k_n'$, $n/\alpha$ cannot be in both disjoint intervals $[k_n,k_n+1)$ and $[k_n',k_n'+1)$, leading to a contradiction.

Now we define $x_n:=n-k_n\alpha$ and $A:=\{x_n\mid n\in\mathbb{Z}\}$. We prove that $A$ is dense in $[0,\alpha]$. Using Problem 2.11, and noting that $1/\alpha$ is also irrational, we deduce that $B:=\{n/\alpha-[n/\alpha]\mid n\in\mathbb{Z}\}$ is dense in $[0,1]$, thus, multiplying by the real number $\alpha$, $C:=\alpha B=\{n-[n/\alpha]\alpha \mid n\in\mathbb{Z}\}$ is dense in $[0,\alpha]$ (given the open interval $(a,b)$ in $[0,\alpha]$ by multiplying by $1/\alpha$ we obtain an interval $(a/\alpha,b/\alpha)$ in $[0,1]$ and thus, being $B$ dense in $[0.1]$, we get some $\gamma\in B\cap(a/\alpha,b/\alpha)$, that multiplied by $\alpha$ becomes $\gamma\alpha\in C\cap(a,b)$, proving the density of $C$ in $[0,\alpha]$).

The last step is to identify $C$ with $A$, that is to identify $[n/\alpha]$ with $k_n$: Both are the greatest integer smaller or equal to $n/\alpha$.